8.10 



APPLIED MECHANIC& 



[COMPRESSION. 



Outer diameter 

 Inner diameter 



72-4 ins. X 72-4-5242 nearly 

 72-0 ina- X 72-0-6184 



Area of ring in circ. ins. 



Cohesive strength of I circ. in. wronght-iron 14600 



Total strength of circumference 846800 Ibs. 



As the area of the end is 4,072 sq. Ins., the boiler 

 would bear a pressure of more than 200 Ibs. per square 

 inch, before parting lengthwise. Indeed, it will be 

 f>.u ud that when uniform thin material is employed, a 

 cylindrical reawl subjected to internal fluid-pressure is 

 almost exactly twice as strong lengthwise as it is circum- 

 ferentially.* When the thickness is considerable, this 

 proportion does not hold good. 



It is a singular fact, that many metals, when drawn 

 into wire, become stronger in respect of tensive strain. 

 This may probably be owing to the wire-drawing process 

 causing the particles to arrange themselves in continuous 

 fibres. From numerous experiments made on iron wire 

 lees than J inch in diameter, it appears that the strength 

 of wire is about one-fourth, or 25 per cent., greater 

 than that of iron bar; so that while the cohesive 

 strength of wrought-iron bars may be taken at 8 tons 

 per square inch of sectional area, that of iron wire 

 may be taken at 10 tons ; or taking bar at 61 tons per 

 circular inch, wire will have a strength of 8 tons per 

 circular inch. 



The following are the rules for computing the strengths 

 of hempen ropes and chains. The dimensions of the ropes 

 are stated in terms of girth or cir- pig. 60. 



cumference, and those of the 

 chains in terms of the diameters 

 of the metal constituting the links. 

 Thus the chain which has a dia- 

 meter of | inch at A (Fig. 60), is 

 called a jf-inch chain. 



We have formed the rules so as 

 to compute the strains at a lower 

 rate than they are generally stated. 

 It may be true that a new chain 

 or rope will bear a much greater 

 strain than that which we have 

 allowed to it ; but the chain loses 

 strength by use not so much 

 from wear as from the particles of 

 iron assuming a crystalline instead 

 of a fibrous texture ; and the rope 

 loses strength by wear as well as 

 by the gradual decay of its fibres. 

 It is preferable, therefore, to err 

 rather on the safe side, especially 

 when it is considered that great 

 damage may often result from the 

 breaking of a chain or rope, and, what is still more care- 

 fully to be guarded against, serious injury to life and 

 limb. 



1. To find the diameter of a chain to carry a given 

 weight. 



Rule. Multiply the weight (tons) to be earned by 

 30 ; the product will be the square of the diameter of 

 the chain reckoned in 16ths of an inch. 



Esample. Required the diameter of a chain to carry 

 10 tons. 10 X 30 = 300, which is near 324, the square 

 of 18 ; therefore the diameter of the chain must be 

 * I inch, or 1} inch. 



2. To find the weight which a chain of given diameter 

 will carry. 



Suit. Divide the square of the diameter (reckoned 



Let r m radina. t = tblekneea, a = eobMtre itrrnftth of 1 aqnare Inch 

 of cuiu, f m preamre to bunt at circumference, and P = preature to 

 tie ea*iif Uofthwtie. 



. ?1 



I u amall compared with r, the aectlonal are* of cailnf U nearly 

 t*rt; and ai fi m ana of end, 



W H P m 1 w rat .: t - J^ = I J. 



in ICths of an inch) by 30 ; the quotient will bo the 

 number of tons carried. 



implc. Required the strength of an inch chain : 

 1 inch is JJ of an inch, and 16 X 16 - 256 ; dividing by 

 30 we have about 8.] tons. 



3. To find the circumference of a rope to carry a 

 given weight. 



Rule. Multiply the weight (tons) by 11 ; the product 

 will be the square of the circumference in inches. 



Example. Required the size of a rope to carry 18 

 tons : 18 X 11 = 198, which is near 196 the square 

 of 14 inches, the circumference required. 



4. To find the weight which a riven rope will carry. 

 R,Ue. Divide the square of the circumference in 



inches byll ; thequotientwillbetheweightcarried in tons. 

 Example. Required the strength of a 4-inch rope : 

 4 X 4 = 16, and 16 divided by 11 gives 1J tons. 



2. COMPRESSION. The strength of materials to 

 resist compressive strains, appears to depend chiefly 

 upon some forces among their particles acting in the 

 opposite direction to cohesive attraction. Indeed, the 

 particles of every solid body appear to be ranged in such 

 positions with respect to one another, and so balanced 

 by cohesive attraction keeping them together on the one 

 side, and by some repulsion resisting their nearer ap- 

 proach on the other, that considerable force is generally 

 required to alter their relative distances. The intensity 

 with which these forces act in any body, appears to be 

 measured by their hardness, or their strength to resist 

 external forces. But while cohesive attraction seems to 

 follow a simple and regular law in any material the 

 amount of attraction or the strength to resist a tensive 

 strain being proportional to the sectional area, or, in 

 other words, to the number of particles upon which the 

 attraction is exerted experiments have as yet shown no 

 very regular law as to strength of materials to resist com- 

 pression. Fortunately, in practice, the other strains to 

 which materials are subjected are generally so much 

 more likely to affect them than mere compressive strain, 

 that when we make the parts of our work sufficiently 

 strong to resist the former, we are tolerably safe in re- 

 apect of the latter. 



Some valuable experiments have been made upon the 

 strengths of building materials wood and iron to 

 resist compression ; and attempts to deduce laws from 

 these experiments have not been wanting. We are not 

 aware, however, that any satisfactory results have at- 

 tended those efforts ; and we fear that any rules founded 

 on our present scanty information in respect of this sub- 

 ject, would tend rather to mislead than to assist the 

 practical mechanic. 



If we suppose that a cubical piece of any material A 

 (Fig. 61) be loaded by a certain weight W, which it is just 

 flg,n, able to bear without 



being crushed, we 

 may readily imagine 

 that a number of such 

 cubes say nine, for 

 instance would bear 

 nine times the weight 

 which the one bears ; 

 and if these cubes 

 were arranged either 

 in a row (Fig. 55) at 

 intervals, or close to- 

 gether, orinasqiiari', 

 they should still all 

 together bear nine 

 times the weight that 

 one of them can bear. When the material we <le.-il with 

 is of uniform consistency, we are therefore warranted in 

 reckoning, that its strength to resist compression, pre- 

 cisely as that to resist tension, is proportional to the 

 area over which the compressive force is spread. Thus, 

 a block of stone 4 feet long and 3 feet broad, should be 

 able to sustain six times the compressing force which 

 a block 2 feet long and 1 foot broad can sustain ; because 

 4x3-12, the area of the one, is six times 2x1-2, 

 the area of the other. 



